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If $P$ partitions a group $G$ and $P \simeq \mathbb{Z}_2^+$, then what can we say about $G$?

Reason: this showed up in trying to prove that if a partition $P$ of a group $G$ is such that for any two elements $A,B \in P$ the product $AB$ is entirely contained in another element $C$, then then one of the elements is a normal subgroup $N$ of $G$ and $P = G/N$. I'm doing this by induction and the $n = 2$ case results in $P$ must be $\mathbb{Z}_2^+$. Well for that case I don't need the result that $P = \mathbb{Z}_2^+$, I later found out, but it still seems like an interesting question, so I posted it here.

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Your general result on partitions can be reformulated in terms of equivalence relations $R$ on $G$ which are compatible with the operation, that is, $a R a'$ and $b R b'$ imply $ab R a'b'$. Then the equivalence class $N = [e]$ of the identity is a normal subgroup, and $[a] = a N$.

(As usual, if $P$ is a partition with your property, then a compatible equivalence relation $R$ is given by $a R a'$ if and only if there is $A \in P$ such that $a, a' \in A$. And if $R$ is a compatible equivalence relation, then $P = \{ [a] : a \in G \}$ is a partition with your property, where $[a] = \{ x \in G : x R a \}$.)

In your particular case when $P \cong \mathbb{Z}_2^+$, you are just considering a group which has a (normal) subgroups of index $2$. The normal subgroup is the element of the partition containing the identity.

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Can we say more than that about the structure of $G$ given that $P$ is isomorphic to $\mathbb{Z}^+_2$? – Enjoys Math Apr 3 '13 at 21:48
@EnjoysMath, no, every group of order $2$ is isomorphic to $\Bbb{Z}_{2}^{+}$. So the condition is just $G$ has a subgroup $N$ of index $2$, and then $N$ is automatically normal. – Andreas Caranti Apr 3 '13 at 21:49
Thanks. I guess what I'm looking for is a comment on what happens when we replace $\mathbb{Z}^+_2$ with a higher order group. – Enjoys Math Apr 3 '13 at 21:52
If you replace $\mathbb{Z}_2^+$ with an arbitrary $H$, say, then you are looking for an extension $1 \to N \to G \to H \to 1$. For any choice of $N$, you have at least the direct product $G = N \times H$, but then you could have many more, depending on $N$ and $H$. See – Andreas Caranti Apr 3 '13 at 22:02

What do you mean by $P$ partitioning $G$? Typically, this language implies that $P$ is a subgroup $G$, and the resulting partition is given by the left (or right) cosets of $P$ in $G$, i.e. $$g_1P,\,g_2P,\, g_3P,\ldots$$

If $G$ is finite, then $P \leq G$ implies that $\# P $ divides $\# G$, i.e. $\# G$ is even. On the other hand, if $\# G$ is even, there exists an element $\sigma \in G$ or order two. Then, taking $P = \langle \sigma \rangle$, we obtain a partition of the form you seek. In other words, if $G$ is finite, then this holds if and only if $\#G$ is even.

We can say less when $G$ is infinite. As above, such a partition exists if and only if there exists an element $\sigma \in G$ of order two. If $G$ is abelian, this holds if and only $G$ has a non-trivial $2$-power torsion subgroup. If $G$ is not abelian, the $2$-power torsion won't necessarily form a subgroup of $G$, and we can say practically nothing about the existence/non-existence of $P$.

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